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Nataliya [291]
2 years ago
11

You may wonder who the rest goes

Engineering
1 answer:
Deffense [45]2 years ago
6 0

Answer:

no not really

Explanation:

From your friendly neighborhood cereal killer,

Sir. BLOODPR1NCE

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Using Von Karman momentum integral equation, find the boundary layer thickness, the displacement thickness, the momentum thickne
Alex_Xolod [135]

Answer:

Explanation:

We can solve Von Karman momentum integral equation as seen below using following in the attached file

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3 years ago
If the load parameters are: Vln=600kV, Il=100A (resistive), calculate the source voltage and current when the line is 50Miles (s
Archy [21]

s 0Miles (short), 150 Miles(medium), and 300 Miles (long).

Explanation:

4 0
3 years ago
When designing a car that runs on wind or Air car . can you tell me the details for the following points Compressed Air Engine:
BabaBlast [244]

Answer:

a)

The crack and connecting rod is used in the design of car.This mechanism is known as slider -crank mechanism.

Components:

1.Inlet tube

2. Wheel

3. Exhaust

4. Engine

5.Air tank

6.Pressure gauge

7.Stand

8. Gate valve

b)

The efficiency of air engine is less as compare to efficiency of electric engine and this is not ecofriendly because it produce green house gases.These gases affect the environment.

c)

it can run around 722 km when it is full charge.

                                                                                                                                                     

5 0
3 years ago
Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0
3 years ago
Outline the process used to test the following hypothesis: Titanium cages are stronger than steel cages for hockey goalie masks.
son4ous [18]

Answer:

true

Explanation:

8 0
3 years ago
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